# 2020 AIME II Problems/Problem 1

## Problem

Find the number of ordered pairs of positive integers $(m,n)$ such that ${m^2n = 20 ^{20}}$.

## Solution

In this problem, we want to find the number of ordered pairs $(m, n)$ such that $m^2n = 20^{20}$. Let $x = m^2$. Therefore, we want two numbers, $x$ and $n$, such that their product is $20^{20}$ and $x$ is a perfect square. Note that there is exactly one valid $n$ for a unique $x$, which is $\tfrac{20^{20}}{x}$. This reduces the problem to finding the number of unique perfect square factors of $20^{20}$. $20^{20} = 2^{40} \cdot 5^{20} = \left(2^2\right)^{20}\cdot\left(5^2\right)^{10}.$ Therefore, the answer is $21 \cdot 11 = \boxed{231}.$

~superagh

~TheBeast5520

## Solution 2 (Official MAA)

Because $20^{20}=2^{40}5^{20}$, if $m^2n = 20^{20}$, there must be nonnegative integers $a$, $b$, $c$, and $d$ such that $m = 2^a5^b$ and $n = 2^c5^d$. Then $$2a + c = 40$$ and $$2b+d = 20$$ The first equation has $21$ solutions corresponding to $a = 0,1,2,\dots,20$, and the second equation has $11$ solutions corresponding to $b = 0,1,2,\dots,10$. Therefore there are a total of $21\cdot11 = 231$ ordered pairs $(m,n)$ such that $m^2n = 20^{20}$.

~ pi_is_3.14

## Video Solution

~ North America Math Contest Go Go Go

~IceMatrix

~avn

## Purple Comet Math Meet April 2020

Notice, that this was the exact same problem (with different wording of course) as Purple Comet HS problem 3 and remembering the answer, put $\boxed{231}$.

~Lopkiloinm

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 